Q#1
A rope 1 cm in diameter breaks if the tension in it exceeds 500 N. The maximum tension that may be given to a similar rope of diameter 2 cm is(a) 500 N
(b) 250 N
(c) 1000 N
(d) 2000 N.
Answer: (d)
The wire breaks when the maximum permissible stress in the cross-section of the wire is attained. Here 500 N gives maximum permissible stress to a 1 cm diameter wire. Since the area of the cross-section of a 2 cm diameter wire will be four times that of the 1 cm diameter wire, its load carrying capacity will also be four times, i.e. 4 x 500 N = 2000 N.
Q#2
The breaking stress of a wire depends on
(a) material of the wire
(b) length of the wire
(c) radius of the wire
(d) shape of the cross-section
Answer: (a)
The stress is force per unit area. The breaking stress of a wire is a property of its material.
Q#3
A wire can sustain the weight of 20 kg before breaking. If the wire is cut into two equal parts, each part can sustain a weight of
(a) 10 kg
(b) 20 kg
(c) 40 kg
(d) 80 kg
Answer: (b)
The breaking stress in each part of the wire will be the same as the original because of the same material. Since the area remains the same, each part can sustain a weight of 20 kg.
Q#4
Two wires A and B are made of same material. The wire A has a length l and diameter r while the wire B has a length 2l and diameter r/2. If the two wires are stretched by the same force, the elongation in A divided by the elongation in B is
(a) 1/8
(b) 1/4
(c) 4
(d) 8.
Answer: (a)
For wires of the same material the elongation is directly proportional to the length and inversely proportional to the cross-sectional area.
Δl/l = F/AY
Δl = Fl/AY {Here F and Y are constant}
The length of A is half of B, for it, elongation of A will be ½ of B and the area of A is four times of B (Since the area is proportional to the square of the radius), for it, the elongation of A will be ¼th of B. For the combined effect of both length and the area, the elongation of A will be ½*¼ =1/8 of B.
Q#5
A wire elongates by 1.0 mm when a load W is hanged from it. If this wire goes over a pulley and two weights, W each are hung at the two ends, the elongation of wire will be
(a) 0.5 m
(b) 1.0 mm
(c) 2.0 mm
(d) 4.0 mm
Answer: (b)
The elongation will be the same because in both cases it is pulled by the same force at its both ends. In the case of hanging wire, the lower end is pulled by the weight W downwards and the upper end is pulled by the support upwards by the same magnitude of the force.
Q#6
A heavy uniform rod is hanging vertically from a fixed support. It is stretched by its own weight. The diameter of the rod is
(a) smallest at the top and gradually increases down the rod
(b) largest at the top and gradually decreases down the rod
(c) uniform everywhere
(d) maximum in the middle.
Answer: (a)
For a hanged heavy uniform rod, at any cross-section, the load is only of the part of the rod which is below it. At its lowest point, the load is zero which gradually increases as we go up. Since the cross-section is the same, the stress increases as we go up. The maximum stress will be at the top and so will be the effect of elongation. Since the transverse strain is proportional to longitudinal strain, hence the diameter of the rod will be smallest at the top and gradually increase down the rod.
Q#7
When a metal wire is stretched by a load, the fractional change in its volume ΔV/V is proportional to
(a) Δl/l
(b) (Δl/l)²
(c) √(Δl/l)
(d) none of these.
Answer: (a)
When a metal wire is stretched by a load the fractional change in the transverse length is proportional to the fractional change in the longitudinal length. i.e.,
Δd/d = σ*Δl/l, where σ is the Poisson's ratio.
Let the original cross-sectional area = A and final area a.
Change in volume ΔV = a(Δl) and the original volume V = Al
Hence the fractional change in the volume
ΔV/V = aΔl/Al = (a/A) (Δl/l) =(d'²/d²)(Δl/l)
={(d – Δd)²/d²}(Δl/l) = {(d² – 2dΔd)/d²}(Δl/l)
[taking Δd² negligible]
So, ΔV/V = {1 – 2Δd/d}(Δl/l)
= {1 – 2σ(Δl/l)}(Δl/l)
= Δl/l – 2σΔl²/l²
ΔV/V = Δl/l {Taking Δl² negligible}
Hence the option (a).
Q#8
The length of a metal wire is l₁ when the tension in it is T₁ and is l₂ when the tension is T₂. The natural length of the wire is
(a) (l₁ + l₂)/2
(b) √(l₁l₂)
(c) (l₁T₂ – l₂T₁)/(T₂ – T₁)
(d) (l₁T₂ + l₂T₁)/(T₂ + T₁)
Answer: (c)
If the original length of the wire be l and the area of the cross-section a, then
l₁ – l = T₁l/Ya
And l₂ – l = T₂l/Ya
Dividing we get,
(l₁ – l)T₂ = (l₂ – l)T₁
l₁T₂ – lT₂= l₂T₁ – T₁l
l(T₂ – T₁) = l₁T₂ – l₂T₁
l = (l₁T₂ – l₂T₁)/(T₂ – T₁)
Hence the option (c).
Q#9
A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break
(a) when the mass is at the highest point.
(b) when the mass is at the lowest point
(c) when the wire is horizontal
(d) at an angle of cos⁻¹(1/3) from the upward vertical.
Answer: (b)
(a) 10 kg
(b) 20 kg
(c) 40 kg
(d) 80 kg
Answer: (b)
The breaking stress in each part of the wire will be the same as the original because of the same material. Since the area remains the same, each part can sustain a weight of 20 kg.
Q#4
Two wires A and B are made of same material. The wire A has a length l and diameter r while the wire B has a length 2l and diameter r/2. If the two wires are stretched by the same force, the elongation in A divided by the elongation in B is
(a) 1/8
(b) 1/4
(c) 4
(d) 8.
Answer: (a)
For wires of the same material the elongation is directly proportional to the length and inversely proportional to the cross-sectional area.
Δl/l = F/AY
Δl = Fl/AY {Here F and Y are constant}
The length of A is half of B, for it, elongation of A will be ½ of B and the area of A is four times of B (Since the area is proportional to the square of the radius), for it, the elongation of A will be ¼th of B. For the combined effect of both length and the area, the elongation of A will be ½*¼ =1/8 of B.
Q#5
A wire elongates by 1.0 mm when a load W is hanged from it. If this wire goes over a pulley and two weights, W each are hung at the two ends, the elongation of wire will be
(a) 0.5 m
(b) 1.0 mm
(c) 2.0 mm
(d) 4.0 mm
Answer: (b)
The elongation will be the same because in both cases it is pulled by the same force at its both ends. In the case of hanging wire, the lower end is pulled by the weight W downwards and the upper end is pulled by the support upwards by the same magnitude of the force.
Q#6
A heavy uniform rod is hanging vertically from a fixed support. It is stretched by its own weight. The diameter of the rod is
(a) smallest at the top and gradually increases down the rod
(b) largest at the top and gradually decreases down the rod
(c) uniform everywhere
(d) maximum in the middle.
Answer: (a)
For a hanged heavy uniform rod, at any cross-section, the load is only of the part of the rod which is below it. At its lowest point, the load is zero which gradually increases as we go up. Since the cross-section is the same, the stress increases as we go up. The maximum stress will be at the top and so will be the effect of elongation. Since the transverse strain is proportional to longitudinal strain, hence the diameter of the rod will be smallest at the top and gradually increase down the rod.
Q#7
When a metal wire is stretched by a load, the fractional change in its volume ΔV/V is proportional to
(a) Δl/l
(b) (Δl/l)²
(c) √(Δl/l)
(d) none of these.
Answer: (a)
When a metal wire is stretched by a load the fractional change in the transverse length is proportional to the fractional change in the longitudinal length. i.e.,
Δd/d = σ*Δl/l, where σ is the Poisson's ratio.
Let the original cross-sectional area = A and final area a.
Change in volume ΔV = a(Δl) and the original volume V = Al
Hence the fractional change in the volume
ΔV/V = aΔl/Al = (a/A) (Δl/l) =(d'²/d²)(Δl/l)
={(d – Δd)²/d²}(Δl/l) = {(d² – 2dΔd)/d²}(Δl/l)
[taking Δd² negligible]
So, ΔV/V = {1 – 2Δd/d}(Δl/l)
= {1 – 2σ(Δl/l)}(Δl/l)
= Δl/l – 2σΔl²/l²
ΔV/V = Δl/l {Taking Δl² negligible}
Hence the option (a).
Q#8
The length of a metal wire is l₁ when the tension in it is T₁ and is l₂ when the tension is T₂. The natural length of the wire is
(a) (l₁ + l₂)/2
(b) √(l₁l₂)
(c) (l₁T₂ – l₂T₁)/(T₂ – T₁)
(d) (l₁T₂ + l₂T₁)/(T₂ + T₁)
Answer: (c)
If the original length of the wire be l and the area of the cross-section a, then
l₁ – l = T₁l/Ya
And l₂ – l = T₂l/Ya
Dividing we get,
(l₁ – l)T₂ = (l₂ – l)T₁
l₁T₂ – lT₂= l₂T₁ – T₁l
l(T₂ – T₁) = l₁T₂ – l₂T₁
l = (l₁T₂ – l₂T₁)/(T₂ – T₁)
Hence the option (c).
Q#9
A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break
(a) when the mass is at the highest point.
(b) when the mass is at the lowest point
(c) when the wire is horizontal
(d) at an angle of cos⁻¹(1/3) from the upward vertical.
Answer: (b)
When the wire is whirled in a vertical circle, the weight is always downward but the centrifugal force is always outward from the center i.e. changing its direction. The resultant of the weight and the centrifugal force will be maximum when both are in the same line and in the same direction. This situation is when the rotating mass is at the lowest point.
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